 Hi and welcome to the session. Let's work out the following question. The question says a body is projected with a velocity of 24 meter per second at an angle of 60 degree with the horizontal. Find first the equation of its path, second its time of flight and third the maximum height attained by it. So let U be the velocity of projection that is 24 meter per second. Let alpha be the angle of projection that is 60 degree. Let us start with the solution to this question. First of all we see that the equation of its path or trajectory is given by y is equal to x tan alpha minus gx square divided by 2u square cos square alpha. This is equal to x tan 60 degree because alpha is 60 degree minus gx square divided by 2 into u square that is 24 square into cos square 60 degree. This is equal to x root 3 minus gx square divided by 2 into 576 multiplied by 1 by 2 the whole square. This is equal to x root 3 minus gx square divided by 288. So the equation will be y is equal to x root 3 minus gx square divided by 288 and this is our answer to first part of the question. Now let us see the second part. Let the time of flight e is given by 2u sin alpha divided by g that is equal to 2 into 24 into sin 60 degree divided by g that is equal to 2 into 24 into root 3 by 2 divided by g is 9.8 meter per second square. So this is equal to 4.24 seconds. So our answer to the second part is time is 4.24 seconds. Now let us see the solution to third part. Here let h be maximum height that is this is h so h is equal to u square sin square alpha divided by 2g. This is equal to 24 square into sin 60 degree the whole square divided by 2 into 9.8 that is equal to 576 into root 3 by 2 the whole square divided by 2 into 9.8. This is equal to 576 into 3 by 4 divided by 2 into 9.8 and that is equal to 22.04 meters. So we have the maximum height attained that is capital H is 22.04 meters. So this is our answer to this question. I hope that you understood the solution and enjoyed this session. Have a good day.